Thread: Inequalities problem

Question: Let a and b be the roots of the quadratic equation $\displaystyle x^2- (m +\frac 3 m )x +2 = 0$ where m is a non-zero constant.
Using the fact that $\displaystyle (a-b)^2 \geq 0$ for all real numbers a and b, show that $\displaystyle m^2 +\frac 9 {m}^2 \geq 6$.
Hence show tha$\displaystyle t (m + \frac 3 m }^2 \geq 12$

MY WORK:
$\displaystyle (a-b)^2 = (a+b)^2 - 4ab = (a+b)^2 - 8 \geq 0$
$\displaystyle (m+ \frac 3 m ) -8 \geq 0$
$\displaystyle m^2+6m+\frac 9 {m}^2 \geq 8 $

Thanks in advance.

Originally Posted by cakeboby

Question: Let a and b be the roots of the quadratic equation $\displaystyle x^2- (m +\frac 3 m )x +2 = 0$ where m is a non-zero constant.
Using the fact that $\displaystyle (a-b)^2 \geq 0$ for all real numbers a and b, show that $\displaystyle m^2 +\frac 9 {m}^2 \geq 6$.
Hence show tha$\displaystyle t (m + \frac 3 m }^2 \geq 12$

MY WORK:
$\displaystyle (a-b)^2 = (a+b)^2 - 4ab = (a+b)^2 - 8 \geq 0$
$\displaystyle (m+ \frac 3 m ) -8 \geq 0$
$\displaystyle m^2+6m+\frac 9 {m}^2 \geq 8 $

Thanks in advance.

From your first line of working onwards, you'll actually want to use

$\displaystyle (m + \frac{9}{m})^2 - 4(m)(\frac{3}{m}) \geq 0$

$\displaystyle m^2 + 6 + (\frac{9}{m})^2 - 12 \geq 0 $

$\displaystyle m^2 + (\frac{9}{m})^2 \geq 6 $

Check the binomial expansion. You did this incorrectly in your working.